Convolution Integral Example 05 - Convolution Of Unit Step With Pulse 

You’re welcome, thanks for watching. Make sure to check out my website adampanagos.org where I have additional resources you might find useful, and consider supporting development of new videos through www.patreon.com/agpanagos. Thanks! Adam

Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content you might find helpful. Thanks, Adam

+Vishakh Kannan Glad to hear that, thanks for watching!

Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (540+ videos) you might find helpful. Thanks, Adam

Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam

+BlastDaMusicUp You're welcome, glad I was able to help!

Glad to help; thanks for watching!

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam

You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.

+Agha Mujtaba Glad to help, I appreciate the nice feedback. Thanks for watching!

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam

Thanks!

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam

Glad to help, thanks!

Glad to hear that, thanks for watching!

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content you might find helpful. Thanks much, Adam

Glad I could help, thanks for watching!

Thanks so much, I appreciate the nice feedback. Best, Adam

You're welcome, hope I was able to help. Thanks, Adam

Glad the video helped, thanks!

Glad I could help, thanks for watching!

Glad I could help, thanks for watching!

Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org where I have a lot of other videos and resources available that you might find helpful. Thanks, Adam.

You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.

For most of my videos, I use Doceri (an app for my iPad, apps.apple.com/us/app/doceri-remote/id412443803?ls=1) that lets me pre-record all the writing. Then, I can "play back" the writing while I record the audio. It’s a very useful tool for me. Hope that helps, Adam

You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.

Thanks for the nice feedback; much appreciated.

Glad to hear that, thanks!

Thanks much! Make sure to check out my website adampanagos.org for additional material you might find helpful.

Glad to help, thanks for watching!

You're welcome, thanks for watching!

+Akash Pandey That's a great question. Fortunately, I think the math works out either way. For Case 1, when t = 1, the edge of y(t-tau) just "barely" touches the start of the unit step function u(tau) at tau = 0. If we assume that there is no overlap (which is what I did in the video), then obviously their product is zero and thus the convolution integral is zero for this case. Lets now consider the case you mentioned. What if we assume that y(t-tau) and u(tau) DO have a non-zero product when t = 1. This assumptions means that we assume u(0) = 1 and y(1) = 1. Note that the overlap occurs only at the SINGLE point tau = 0. So, while the signals do overlap and thus have a product equal to 1, they only have this value for the single point tau = 0. Thus, when integrating you would have int_{0}^{0}1 dtau which by definition is zero since you're integrating over a width of zero. So, even if you assume that they do overlap at this singular point you'll still end up with an integral of zero. Does that help a little?

For most of my videos, I use Doceri (an app for my iPad, apps.apple.com/us/app/doceri-remote/id412443803?ls=1) that lets me pre-record all the writing. Then, I can "play back" the writing while I record the audio. It’s a very useful tool for me. Hope that helps, Adam

Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam

its really urgent

You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.

For most of my videos, I use Doceri (an app for my iPad, apps.apple.com/us/app/doceri-remote/id412443803?ls=1) that lets me pre-record all the writing. Then, I can "play back" the writing while I record the audio. It’s a very useful tool for me. Hope that helps, Adam

+Vinh Dinh Glad to hear that, thanks for watching!

You're welcome, thanks for watching!

If you're not sure where tau comes from, you probably need to go review the definition of a convolution integral before diving into worked examples. See the definition here:en.wikipedia.org/wiki/Convolution

I'm not sure I understand your comment. The signal x(tau) is zero for all tau < 0. It "turns on" at tau = 0. So, the first case we consider for y(t-tau) is when its front edge, is t-1 < 0. I'm not sure why you think it's for when t-1 < 1? Can you provide some more details perhaps?

But the signal x(tau) begins from 1............... So there will be no overlapping of x(tau) and y(t-tau) till when t-1

No, you're looking at the wrong signal. x(tau) = u(tau). u(tau) is the unit step function which turns on at time tau = 0. We are convolving x(t) with y(t). x(t) starts at time t = 0. y(t) starts at time t = 1. There are two different signals in this problem.

There is no conceptual error but just a human error.Hope you get me this time

Oh sir! I made a blunder.......sorry for killing your time.....

Yes, we're all living in the covid nightmare. Hang in there. Hope all is well. Best, Adam

Good point. Was not explained.